Problem 1:Moving with charge
A proton moves in the same direction as the current in an infinite wire. For constant current the Lorentz force on the charge is toward the wire. Now go to a reference frame which moves with the proton. In this frame the proton is not moving. Therefore the proton experiences no force toward the wire.
Not strictly true. If you had a conducting wire and a charge, we could easily use the method of images to find the voltage and take that result and find a force on the proton. There is an attractive electric force here. I hate to introduce special relativity to the discussion, but magnetism is a simplification as it drops out of electric force with relativistic treatment of its field. The magnetic force that we calculate in the "stationary" frame is just an easy way to calculate the relativistic electric force.
The assumption in this problem that the velocity, v, is frame dependent is not valid. It has to be in reference to the wire. This is very similiar to a friction problem that is also velocity dependent. If an object is sliding on a surface, it cannot be suggested that there would be a different force in the object's reference frame because v=0. The object's velocity must be referred to the surface it is sliding over, or in this case to the wire.
In reference to the second paragraph, how can we know if we are in a "stationary frame"? I think that to test this out, we should take a wire and a proton, hold them out in front of us, drive down the highway, and measure the attraction between the two. Even though this is not a "stationary frame" the Lorentz force is not apparent. I agree that the attraction is reference frame dependent. It seems as though we can only observe the Lorentz force if we are in a different reference frame than the "moving" charges. Not that this makes a whole lot of sense....but what does?
We know we're in a "stationary frame" when we experience no acceleration. There will be a trade-off between magnetic and electric forces depending on the frame we choose, but the net force is "coincidentally" the same from every inertial frame. The net Force is what dictates motion. If we got different motions calculating from different frames, we'd have a problem. The fact that we have a velocity dependent force indicates that the magnetic and electric forces must change depending on the frame we choose to calculate from. All inertial frames are valid.
I think that the assumption made in the final part of the proton with no net force cannot be true. At any instantaneous moment, the Lorentz force of qv x B will be acting to pull the electron toward the wire. The inertial frame of the wire would have to be continually accelerating to counter this because, due to the changing direction of B at different points around the wire, the Lorentz force is constantly changing in magnitude and direction. I guess, this could technically be said to be equal to no net force, but the frame of the wire would have to have an outside force equal in magnitude and direction to the Lorentz force acting on the proton.
Why not just use the pre-existing Lorentz Force equation qv x B but instead of having v being the velocity of the proton (in the proton reference frame) have it be the velocity of the wire compared to the proton, which would be equal to the v in a stationary reference frame just opposite direction. This will change the direction of the Lorentz Force but it's ok because now you are viewing the wire being attracted to the proton rather than the proton to the wire.
Initially I thought that if you stopped the reference frame you would have a repulsive force between the charge and what becomes a linear charge density. Further reading indicates that this is not what is happening. First, the wire is actually neutral, therefore if the wire is stopped there is no repulsive or attractive force between the wire and charge (this is disregarding the image case as referenced above. We do have a negative current flowing in the opposite direction as the previous part and so we have essentially the same magnetic field since charge and direction are both negated. What is bothering me is that the v vector for the force on the proton seems to be zero which would imply a zero force but this is not the case.
